Selecting at least one ball of each color An urn contains five red, six white and seven blue balls. Five balls are selected without replacement. Find the probability that at least one ball of each color is selected.
Answer (attempt):
Getting red balls = $5 \choose 1$
Getting white balls = $6 \choose 1$
Getting blue balls = $7 \choose 1$
Remaining two balls = $15 \choose 2$
Total selections = $18 \choose 5$
P = $\frac{{5 \choose 1} {6 \choose 1}{7 \choose 1}{15 \choose 2}}{18 \choose 5}$
The answer seems to be wrong.
 A: The total number of combinations is:
$$\dbinom{5+6+7}{5}=\dfrac{18!}{5!\cdot13!}=8568$$

The number of combinations with no red balls is:
$$\dbinom{6+7}{5}=\dfrac{13!}{5!\cdot8!}=1287$$

The number of combinations with no white balls is:
$$\dbinom{5+7}{5}=\dfrac{12!}{5!\cdot7!}=792$$

The number of combinations with no blue balls is:
$$\dbinom{5+6}{5}=\dfrac{11!}{5!\cdot6!}=462$$

The number of combinations with no red balls and no white balls is:
$$\dbinom{7}{5}=\dfrac{7!}{5!\cdot2!}=21$$

The number of combinations with no red balls and no blue balls is:
$$\dbinom{6}{5}=\dfrac{6!}{5!\cdot1!}=6$$

The number of combinations with no white balls and no blue balls is:
$$\dbinom{5}{5}=\dfrac{5!}{5!\cdot0!}=1$$

So the probability of a combination without at least one ball of each color is:
$$\dfrac{1287+792+462-21-6-1}{8568}=\dfrac{2513}{8568}$$

And the probability of a combination with at least one ball of each color is:
$$1-\dfrac{2513}{8568}=\dfrac{6055}{8568}\approx0.7067$$
A: For the split up of $5$ there are $2$ possibilities:
$5=1+2+2$
$5=1+1+3$
Leading to $$\binom{5}{1}\binom{6}{2}\binom{7}{2}+\binom{5}{2}\binom{6}{1}\binom{7}{2}+\binom{5}{2}\binom{6}{2}\binom{7}{1}+\binom{5}{1}\binom{6}{1}\binom{7}{3}+\binom{5}{1}\binom{6}{3}\binom{7}{1}+\binom{5}{3}\binom{6}{1}\binom{7}{1}$$
possibilities under the condition that from each color at least one ball is selected. 
In total there are: $$\binom{18}{5}$$ possibilities. 
These facts enable you to find the probability.
